Canon makes some really awesome DSLR cameras. What they don't
do is tell you how to get the most out of your batteries. Sure,
there are general rules like turn the camera off when not in
use, keep the batteries fully charged, etc. - but to the best of my
knowledge, to date nobody has actually quantified how much
various features really cost in terms of battery life.

Until now!

This all started because a newly acquired EOS 50D for my wife seemed to eat batteries for breakfast,
lunch, and dinner. My first thought was this might be a
recurrence of a problem I'd previously had with my first EOS
650. So I recreated the same tool I'd done back in 1987 now for
the new DSLRs, and took measurements. My goal is to translate
Electrical Engineer-Speak into some real world numbers that
everyone can understand.

I'm going to step out of an engineering mindset and not
be 100% accurate, but "good enough" for this discussion. In other words,
I'm sacrificing some accuracy and precision to error on the side of
readability.

If you look at a rechargeable battery, there is a
specification on the outside for its capacity, specified in Amp-Hours
(Ah) or milliamp-hours (mAh). My Canon BP-511A battery part number
E160814 says "7.4V 1390 mAh"

The most common analogy for electricity is water - if
you put water in a jug and run a tube out of the jug so water flows
through the tube, the water itself is the current, measured in Amps and
the pressure pushing that water through the tube is the voltage,
measured in Volts. Together, they can do work, measured in watts, and
for batteries the watts of power is the product of the volts * the amps
of capacity. To make things more straightforward, I'm ignoring watts and
the conversions throughout this article.

So the water in our jug - the current - is our capacity.
As you start using the water, it drains until there is nothing left, at
which point your battery is dead (and so is your camera!).

1390 mAh means you can draw 1390 milliamps for one hour
before the battery is empty.

Or you can draw half that (1390/2 = 695 mA) for two
hours before the battery is empty.

Or you can draw 1/10th of the capacity, 139 milliamps
for 10 hours before the battery is empty.

On the other side, you can draw twice the current, or
1390 * 2 = 2780 milliamps for 1/2 hour and have an empty battery.

To very briefly touch some of the subtleties, battery
capacity is specified for a new battery - they degrade with time and
charge / discharge cycles, so your 2 year old battery won't have the
same capacity as a new one. The capacity is specified at a specific
temperature, so if your battery is cold it won't have the same ability
to give up all its current to the camera. If you've left the battery in
a hot car every day over the summer, it has also reduced its capacity.
In other words, I'm talking about a best case scenario here.

The last bit of data here is that 1 Amp = 1000 milliamps.
In electrical engineering, metric system prefixes are everywhere.

Enough
with the engineering! So what did you do?

If you read the other article, I did the same thing I did then only for
a modern Canon EOS DSLR. Then I measured various characteristics to
determine how much current is consumed by the camera's various features.
This way, when you use a feature you can have in the back of your mind
"This is how much of my battery this feature is consuming per minute,
shot, or some other measure." With that knowledge, you can possibly
decide to change how you use your camera and get the most out of it by
turning on or off features based on what you have for battery capacity.

This shows my test rig.
Starting with a 3rd party BP511 clone battery on the left with the white
and purple label, I opened it up, removed the batteries, and attached wires to the connectors. Once done, this
became a "Dummy battery" that was used to bring the internal camera
connections out. Once connected to a real BP-511a battery, I could then
power the camera and measure how much current the camera was consuming
when the camera was doing various tasks.

Results!

I measured all kinds of things - Image Stabilization on a few different lenses, taking a picture, servo mode and
the camera constantly adjusting the focus, the draw of different types of autofocus motors (USM, the older AFD in my EF 35-70mm lens, ...)
... more than you can imagine.

If I get enough requests, I'll be happy to publish raw
data.

But the biggest single thing you can do to get the most
shots out of a single battery is this:

KEEP THE LCD OFF AS MUCH AS POSSIBLE!

When the LCD is lit, that is the biggest continuous
drain your camera will place on your batteries.

The biggest current spike is actually taking a picture.
Those motors that lift the mirror, open the shutter, and move the mirror
back down again suck a big spike. But it is only for a fraction of a
second, so it isn't that bad. But it does need a big spike - that is why
a marginal battery showing one bar on the display might turn into an
instantly empty battery flashing icon on the display when you try to
take a picture. The high current draw of the motors can't get enough
juice, the camera detects this, and what you thought was a half-full
battery turns into an empty battery.

The built-in flash is also a big draw, but that drops
once the flash is charged.

But that LCD, the folks that chimp on it and review
photos a lot, those are the cases where your battery never seems to last
as many shots as you hoped.

In our case, the feature of the EOS 50D where the back
LCD display shows all the functions of the camera and is on all the time
will suck the batteries dry faster than any other "feature". I didn't
measure the EOS Rebel cameras, but I imagine the results would be
similar with those as that power has to come from somewhere.